# Bartle 6.K elements of integration

I have a question about an equality of the expoents we are asked to show.

The book states: $$||f||_r <= ||f||_p * \mu (X)^s$$ where $$s=1/r - 1/p$$ when $$\mu (X) < \infty$$

The thing is when using holder inequation im getting $$\mu (X)^{\dfrac{1}{r-p}}$$ and its driving me crazy that i can't show $${\dfrac{1}{r-p}} =s$$

The only relation i have is the one i've used for holder's inequation, and its $$p/r + (r-p)/r = 1$$

• You didn't just write $\dfrac 1r - \dfrac 1s = \dfrac1{r-p}$, did you? – Ted Shifrin Jun 14 at 23:38
• Nop, i have done some Dumb maths, Just don't know where – Daniel Moraes Jun 15 at 10:54

$$\int |f|^{r} \leq (\int |f|^{p})^{r/p}(\mu(X))^{1-r/p}$$ so $$\|f||_r \leq\|f\|_p (\mu(X))^{s}$$. I have applied Holder's inequality with indices $$\frac p r$$ and $$\frac p {p-r}$$.